Solve for $x$ and $y$ using substitution. ${x+3y = 5}$ ${x = 2y-10}$
Since $x$ has already been solved for, substitute $2y-10$ for $x$ in the first equation. ${(2y-10)}{+ 3y = 5}$ Simplify and solve for $y$ $2y-10 + 3y = 5$ $5y-10 = 5$ $5y-10{+10} = 5{+10}$ $5y = 15$ $\dfrac{5y}{{5}} = \dfrac{15}{{5}}$ ${y = 3}$ Now that you know ${y = 3}$ , plug it back into $\thinspace {x = 2y-10}\thinspace$ to find $x$ ${x = 2}{(3)}{ - 10}$ $x = 6 - 10$ ${x = -4}$ You can also plug ${y = 3}$ into $\thinspace {x+3y = 5}\thinspace$ and get the same answer for $x$ : ${x + 3}{(3)}{= 5}$ ${x = -4}$